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I have to prove (if $\gamma \ne 0$) that there is a analytic continuation for $\Re s >0$ of the function $$f(s)=\frac{\zeta (s)^2 \zeta(s-i\gamma )\zeta(s+i\gamma ) }{\zeta(2s)} $$ and that this continuation has possible singularities at $s=1,~s=1\pm i\gamma$ and $\Re s =1/2$.

The analytic continuation is clear since $\zeta $ has such a continuation - is this sufficient? And concerning the singularities: $s=1,~s=1\pm i\gamma$ is obvious since $\zeta$ has a pole at $s=1$. But what about $\Re s =1/2$?

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  • $\begingroup$ The assertion seems wrong: there are no singularities on $\Re s=\frac12$, and the zeros $\rho$ of $\zeta(2s)$ with positive real part all satisfy $\Re \rho<\frac12$ and, assuming the Riemann hypothesis, $\Re \rho=\frac14$. $\endgroup$ – Greg Martin Oct 14 '14 at 21:20
  • $\begingroup$ Okay, yes that's even what I thought. The next step I should prove is that it has a removable singularity at the real point $s=1/2$. Is this just because the numerator is bounded and the denominator diverges? $\endgroup$ – sBs Oct 14 '14 at 21:44
  • $\begingroup$ Essentially, yes. $\zeta(s)$ has a simple pole with at $s=1$, which means, by definition, that $\zeta(s)(s-1)$ has a removable singularity. Hence $1/(2s-1)\zeta(2s)$ also has a removable singularity. $\endgroup$ – Greg Martin Oct 14 '14 at 23:31

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